Question

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times {10}^{3}kg{m}^{-3}$ ?

Answer 1

Let the given depth be $h$.
Pressure at the given depth, $p=80.0atm=80\times 1.01\times {10}^{5}Pa$
Density of water at the surface, ${\rho }_1=1.03\times {10}^{3}kg/m^{-3}$
Let ${\rho }_2$ be the density of water at the depth $h$.
Let $V_1$ be the volume of water of mass $m$ at the surface.
Let $V_2$ be the volume of water of mass $m$ at the depth $h$.
Let $△V$ be the change in volume.
$△V=V_1-V_2$
$=m\left[\right(1/{\rho }_1)-(1/{\rho }_2\left)\right]$

$\therefore$ Volumetric strain $=△V/V_1$
$=m\left[\right(1/{\rho }_1)-(1/{\rho }_2\left)\right]\times \left({\rho }_1/m\right)$

$△V/V_1=1-\left({\rho }_1/{\rho }_2\right)$ .....(i)
Bulk modulus, $B=p{V}_1/△V$
$△V/V_1=p/B$
Compressibility of water $=\left(1/B\right)=45.8\times {10}^{-11}P{a}^{-1}$

$\therefore △V/V_1=80\times 1.013\times {10}^5\times 45.8\times {10}^{-11}$

$=3.71\times {10}^{-3}$ .....(ii)
For equations (i) and (ii), we get:
$1-\left({\rho }_1/{\rho }_2\right)=3.71\times {10}^{-3}$

${\rho }_2=1.03\times {10}^3/[1-(3.71\times {10}^{-3}\left)\right]$

$=1.034\times {10}^{3}kg{m}^{-3}$
Therefore, the density of water at the given depth $\left(h\right)$ is $1.034\times {10}^{3}kg{m}^{-3}$.